$l^p (1≤p<{\infty})$ and $L^p (1≤p<∞)$ are separable spaces.

What on earth has changed when the value of $p$ turns from a finite number to ${\infty}$?

Our teacher gave us some hints that there exists an uncountable subset such that the distance of any two elements in it is no less than some $\delta>0$.

Actually I don't understand the question very well, but I hope I have made the question clear enough.

Thank you in advance.

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    $\begingroup$ When you say that you don't understand the question very well, which part is unclear? Do you know what it means for a metric space to be separable, or inseparable? Do you know the definition of the $\ell^\infty$-norm? $\endgroup$
    – user16299
    Jan 9, 2012 at 16:11
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    $\begingroup$ @Andylang: if you think Yemon Choi's comment is good, how about answering some of the clarifying questions he asked? $\endgroup$ Jan 9, 2012 at 16:20
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    $\begingroup$ @ Henning Makholm :Yes,I know the definition of inseparable space.And the norm of $l^p$ (p is finite) is clear for me.I will google the norm of $l^{\infty}$.Thanks. $\endgroup$
    – Andylang
    Jan 9, 2012 at 16:23
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    $\begingroup$ Here's a thread containing the answer math.stackexchange.com/questions/33044/… (I earlier voted to close as a duplicate but given that the closed question is closed too, I'm a bit unsure if closure is a good idea). $\endgroup$
    – t.b.
    Jan 9, 2012 at 16:29
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    $\begingroup$ For $\ell^{\infty}$, you can see here: math.stackexchange.com/questions/170068/… $\endgroup$
    – Seirios
    Feb 15, 2013 at 0:02

2 Answers 2


To be separable means to have a countable dense subset. Suppose that $(M, d)$ is a metric space and that $U\subseteq M$ be an uncountable subset and $r > 0$. Suppose that for all $x \neq y \in U$, $d(x,y) \ge r$. Let $C$ be any countable subset of $M$. Then $C$ can only meet a countable number of the balls $B_{r/2}(x)$ for $x\in U$. Let $G$ be the union of all $B_{r/2}(x)$ for $x\in U$ that do not meet $C$. $G$ is a nonempty open subset of $M$ that does not meet $C$.

There can be no countable dense subset of $M$.

Consider the case of $\ell^\infty$. For each subset $Q$ of the integers, let $x_Q$ be the sequence that is 1 on $G$ and 0 off of it. The $x_Q$ are uncountable and any two elements of this collection are distance 1 apart. We have just shown that $\ell^\infty$ is not separable.

You can generate a similar construct for $L^\infty$. Consider the uncountable subclass of characteristic functions $\{\chi_{B_r(0)}\}_{r>0}\subseteq L^\infty(\mathbb{R}^n)$. Then each pair of distinct elements in it would be 1 unit distance apart. Ergo there cannot be any countable subset of $L^\infty(\mathbb{R}^n)$ that is dense in it.

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    $\begingroup$ (+1) Why do we have to let $G$ be the union of all $B_{r/2}(x)$ that do not meet $C$? Isn't choosing one such $B_{r/2}(x)$ that does not meet $C$ suffice to show that $C$ is not dense? (using this definition: A is dense in X if for any point x in X, any neighborhood of x contains at least one point from A) $\endgroup$
    – yoyostein
    Nov 14, 2016 at 16:03
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    $\begingroup$ Thank you ncmathsadist and I would like to see an answer to the comment of @yoyostein, thanks again $\endgroup$
    – ahdahmani
    Jul 12, 2019 at 20:49

With the topology $\|f-g\|_{l^\infty}=\sup_i |f_i-g_i|$ it is easy to show if C is any countable set, and $(f_{ij})_j$ a sequence of elements of $C$, you can take an element $h\in l^\infty$ so that $h_i= 0$ if $|f_{ii}|>1$, $h_i=2$ otherwise, hence $\|f_{j}-h\|>1/2$ for all $j$. Then no ball of radius $1/2$ around $h$ can contain an element of $C$.

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    $\begingroup$ I think it should be $\|f_j-h\|\geq\|f_{jj}-h_j\|\geq1$ (which, of course, doesn't change the result). $\endgroup$ Jul 21, 2015 at 9:56
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    $\begingroup$ Also "no ball" is a bit misleading since there is just one ball of radius 1/2 around $h$. May "no ball of radius smaller 1" would fit here better. $\endgroup$ Jul 21, 2015 at 9:58
  • $\begingroup$ Why do you say topology? Isn't that just the norm? $\endgroup$ Apr 30, 2019 at 14:03
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    $\begingroup$ The norm induces the topology. $\endgroup$ Apr 30, 2019 at 19:33

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